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(C) In an $LC$ circuit,the total energy $E$ is constant and oscillates between the electric field of the capacitor and the magnetic field of the induc

Vedclass · Accepted Answer

$\frac{\pi}{4} \sqrt{L C}$

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(C) In an $LC$ circuit,the total energy $E$ is constant and oscillates between the electric field of the capacitor and the magnetic field of the inductor.The energy stored in the capacitor is $U_E = \frac{q^2}{2C}$ and the energy stored in the inductor is $U_B = \frac{1}{2}LI^2$.At $t=0$,the charge is $q_0$,so the total energy is $E = \frac{q_0^2}{2C}$.We want the time $t$ when $U_E = U_B = \frac{E}{2}$.Since $U_E = \frac{q^2}{2C} = \frac{1}{2} \left( \frac{q_0^2}{2C} \right)$,we have $q^2 = \frac{q_0^2}{2}$,which implies $q = \frac{q_0}{\sqrt{2}}$.Given $q = q_0 \cos(\omega t)$,we set $q_0 \cos(\omega t) = \frac{q_0}{\sqrt{2}}$,so $\cos(\omega t) = \frac{1}{\sqrt{2}}$.This gives $\omega t = \frac{\pi}{4}$.Since $\omega = \frac{1}{\sqrt{LC}}$,we have $t = \frac{\pi}{4\omega} = \frac{\pi}{4} \sqrt{LC}$.

$A$ fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self-inductance $L$ at $t=0$. The time at which the energy is stored equally between the electric and the magnetic field is

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